What is the solution for y^'+2x^2y=x^2 ?

The solution for y^'+2x^2y=x^2 is y=(e^{(-2x^3+c_{1})/3})/(2)+1/2

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y^'+2x^2y=x^2

y^{^{'}} + 2 x^{2} y = x^{2}

y = \frac{e ^{\frac{- 2 x ^{3} + c _{1}}{3}}}{2} + \frac{1}{2}

Solution steps

y^{'} (x) + 2 x^{2} y = x^{2}

Solve separable ODE:

y = \frac{e ^{\frac{- 2 x ^{3} + c _{1}}{3}}}{2} + \frac{1}{2}

y = \frac{e ^{\frac{- 2 x ^{3} + c _{1}}{3}}}{2} + \frac{1}{2}

\int \frac{2 - 2 x}{1 - x} d x

\frac{d y}{d x} + \frac{y}{x} = 2 x^{8} y^{2}

\frac{\partial}{\partial x} (arctan (9 x))

y = (4 x^{2} + 3) (2 x - 3)

\int 18 arctan (x) d x

What is the solution for y^'+2x^2y=x^2 ?
The solution for y^'+2x^2y=x^2 is y=(e^{(-2x^3+c_{1})/3})/(2)+1/2